酸鹼中和的問題?

這是緩衝溶液的計算，而不是酸鹼中和。

CH3COOH = CH3COO^- + H^+ (Ka)

由於 [CH3COOH] 和 [CH3COO^-] 的濃度足夠高，所以 [H^+] 的改變，對以上兩個濃度無多大改變。
所以 [CH3COOH]eqm = [CH3COOH]o = 0.10 M
而 [CH3COO^-]eqm = [CH3COO^-]o = 0.10 M

平衡時：
Ka = (0.1)[H^+]/(0.1) = 1.8 x 10^-5 M
所以 [H^+] = 1.8 x 10^-5 M
pH = -log[H^+] = 4.74

2006-12-09 03:52:06 補充：
有人文不對題，copy and paste，不懂計算，卻抄錄理論。可恨在公眾投票中，「胡言亂語」者只要抄錄夠長，卻每每鯉躍龍門，頓成「最佳答案」。烏呼哀哉！

(~7)

pH值取決於溶液中酸(H+ / H3O+) 或/及鹼(OH-)的濃度
H+ / H3O+的濃度越高，pH值越高
OH-的濃度越高，pH值越低

Principle of Buffer Action

Consider the CH3COOH / CH3COONa buffer system:

CH3COOH (aq) H+ (aq) + CH3COO- (aq) Ka = 1.8 x 10-5

The CH3COOH/CH3COONa buffer system contains largely:

undissociated CH3COOH (aq)
CH3COO- (aq), the conjugate base
The addition of CH3COO- from CH3COONa suppresses the dissociation of CH3COOH.

If a small amount of acid is added to this system, the added hydrogen ions will be consumed by the large quantity of the conjugate acid, CH3COO- ions, to form the undissociated CH3COOH, thereby removing all the added H+ (aq) added.

CH3COO- (aq) + H+ (aq) CH3COOH (aq)

The result is such that the pH of the buffer remains nearly the same as before.

Likewise, if a small amount of base is added, the undissociated CH3COOH will consume the added OH- (aq) and CH3COO- (aq) are formed.

OH- (aq) + CH3COOH (aq) CH3COO- (aq) + H2O (l)

Again, the result is such that the pH of the buffer remains nearly the same as before.

2006-12-08 10:48:51 補充：
http://www.all4you.com.tw/web3/vip/tips/026_c/main.asp?right1=2

2006-12-24 14:28:09 補充：
有人文不對題，copy and paste，不懂計算，卻抄錄理論。可恨在公眾投票中，「胡言亂語」者只要抄錄夠長，卻每每鯉躍龍門，頓成「最佳答案」。烏呼哀哉！